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be of such a length and placed in such a position

E

Fig. 366.

that the plan may not pass the edges of the paper. Fix a needle at m, and place the table so that the point m, may be over the point M, and the line m n may have as nearly as possible the direction M N. Fix the feet of the stand firmly in the soil, and see that the table is quite horizontal, and that when the cross-bar is placed in the direction mn, on looking through the slit, a staff placed at N may be seen. Fix the table in this position. Taking care not to touch the table, turn the cross-bar round the needle fixed at m, until the line of sight passes through the staff fixed at A; draw along the edge of the ruler the line, ma; the angle amn will be equal to AM N. Direct in succession the cross-bar to all the points marked B, C, D, etc., and in each case draw mb, mc, m d, etc. From time to time make certain that the table has not been moved; that is to say, that mn has exactly the direction M N.

Now, take up the instrument, and fix it at the other station N. Repeat the same operations at this point, and draw the lines na, nb, nc, etc., the angle anm will be equal to the angle A N M, etc. Compare now the figure thus obtained with the polygon to be laid down. The triangles AN M, anm, are similar to each other because they are equiangular, and it is the same with all the others, the ratio of similarity being that of M N to mn. It follows that the ratios of MA, M B, M C, to ma, mb, m c, are equal, and the polygon abcd is similar to the polygon A B CD.

The perimeters of similar polygons have the same ratio as their homologous sides.

341. For since the ratio of each side to the side homologous to it is the same, the ratio of the sum of sides of the first polygon to the sum of the sides of the second is equal to the ratio of any two homologous sides. If, for example, each side of one is three times the homologous side of the other, the perimeter of the first is three times the perimeter of the second. The surfaces of similar polygons have the same ratio as the squares of the homologous sides.

342. We have already seen (§ 281) that the ratio of the surfaces of two similar triangles is equal to the ratio of the squares of the homologous sides. But, two similar polygons may be decomposed into the same number of triangles similar each to each, and having the same ratio of similarity; therefore, as the ratio of the surfaces of two homologous triangles is equal to the ratio of the squares of their homologous sides, the same ratio must exist between the sum of the triangles which form the first polygon and the sum of those which form the second.

Circumferences have the same ratio to one another as their radii. Circles have the same ratio to one another as the squares of their radii.

343. Suppose that two circumferences are each divided into the same number of equal parts, sufficiently numerous to be very small compared with the radius, two similar polygons will be formed. The greater the number of sides in the polygons, the less will be the difference between their perimeters and the circumferences, and the less the difference between their surfaces and the circles. However great the number, the ratio of the perimeters of the two polygons remains always equal to the ratio of their radii,

and therefore the ratio of the two circumferences is equal to the ratio of their radii. For example :if the radius of a circle is double that of another, then the circumference of the first is double that of the second, but the area of the first is four times that of the second.

To describe a circumference equal to the sum or difference of two others.

344. Since circumferences are proportional to their radii, if the radius of one of them be three yards and that of the other five yards, the first is equal to three times a circumference of one yard radius, and the second five times the same. The circumference which shall be the sum of the two others, will be equal to eight times the circumference of one yard radius, its radius will therefore be eight times a radius of one yard; that is to say, it will be equal to the sum of the radii.

Fig. 367.

We easily see from this proposition that all the lines of the figure 367 have the same length between the points A and B.

To construct a square which shall have to a given square a ratio expressed by a given fraction,— two-thirds, three-fourths, for example.

345. Since two squares are similar figures, the ratio of their surfaces is equal to the ratio of the squares of their sides. We may therefore find the side of a square, the surface of which has a given ratio to that of a given square, if we can draw two lines such that the squares of the numbers which measure them may have a given ratio.

Let it be required, for example, to construct a square

which may be two-thirds of another square of which

the side a is given. Take

(fig. 368) any distance, and

mark it off three times

D

from A to B and twice from B to C upon a line AC. Upon A C describe a semicircumference, and at B erect a perpendicular,

meeting the circumference in D;

a

B

Fig. 368.

join AD, DC; take DA' equal to a, and draw A'C' parallel to AC; DC' will be the side of a square two-thirds of the square constructed upon a (§ 273). We have

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Now, the first ratio is that of the surfaces of the required and the given square; therefore D C' is the side of the square required.

We have seen (§ 271) that the three sides B C, B A, CA, of a right-angled triangle, are the hypothenuses of three similar right-angled triangles, A B C, D B A, D C A, of which one, A B C, is the sum of the other two; since the surfaces of the triangles are to one another as the squares of their homologous sides. We see that if we construct three similar polygons upon the three hypothenuses they will be to one another as the triangles, and the greatest of them will be the sum of the other two (fig. 369).

Let a, b, c, d, be the homologous sides of four similar polygons (fig. 370). Draw consecutively rightangled triangles OA B, O B C, O C D, in which OA is equal to a, A B equal to b, B C equal to c, C D equal to d; O B will be side of a polygon

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